(B) Given equation: $2\cos^2 \theta - 2\sin^2 \theta = 1$Take $2$ as a common factor: $2(\cos^2 \theta - \sin^2 \theta) = 1$Using the trigonometric id

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(B) Given equation: $2\cos^2 \theta - 2\sin^2 \theta = 1$Take $2$ as a common factor: $2(\cos^2 \theta - \sin^2 \theta) = 1$Using the trigonometric identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$,we get:$2\cos 2\theta = 1$$\cos 2\theta = \frac{1}{2}$Since $\cos 60^\circ = \frac{1}{2}$,we have:$2\theta = 60^\circ$$\theta = 30^\circ$

Competitive Quantitative Aptitude Trigonometry Trigonometry

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$2\cos^2 \theta - 2\sin^2 \theta = 1$,then $\theta = \dots \dots ^\circ$

A
$15$
B
$30$
C
$45$
D
$60$

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$2\cos^2 \theta - 2\sin^2 \theta = 1$,then $\theta = \dots \dots ^\circ$

Similar Questions

If $2 \sin \left(\frac{\pi}{2}\right) = x^{2} + \frac{1}{x^{2}}$,then the value of $\left(x - \frac{1}{x}\right)$ is

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